To find a number in the given series with the key that, they are unique at some level.
Given f(1) = 1, f(2) = 1, f(3) = 2, f(4) = 3, f(5) = 5, f(6) = 8, f(7) = ? unique at quintic function.
Solution:
This can be written as
n f(n)
1 1
2 1
3 2
4 3
5 5
6 8
7 ?
Calculate the first derivative of the function
n f(n) D1(n)
1 a 1 (b-a) (1-1) = 0
2 b 1 (2-1) = 1
3 2 (3-2) = 1
4 3 (5-3) = 2
5 5 (8-5) = 3
6 8 (?-8) = x
7 ?
Similarly, second, third, fourth and fifth derivatives are calculated
n f(n) D1(n) D2(n) D3(n) D4(n) D5(n)
1 1 (1-1) = 0 (1-0) = 1 (0-1) = -1 (1-(-1)) = 2 (-1-2) = -3
2 1 (2-1) = 1 (1-1) = 0 (1-0) = 1 (0-1) = -1 (y-(-1)) = z
3 2 (3-2) = 1 (2-1) = 1 (1-1) = 0 (x-0) = y
4 3 (5-3) = 2 (3-2) = 1 (w-1) = x
5 5 (8-5) = 3 (v-3) = w
6 8 (?-8) = v
7 ?
Given that, it is unique at quintic function and quintinc is power of 5.
And so the value of fifth derivative will be same.
Therefore, z = -3
and it is known that,
(y-(-1)) = z à refer fifth derivative
substituting the value of z,
y+1 = -3
y = -4
And x-0 = y
that is, x-0 = -4
x = -4
Further,
w-1 = x
now, w-1 = -4
w = -3
Also, v-3 = w
on substituting, v-3 = -3
v = 0
At last,
?-8 = v
which means, ?-8 = 0
and therefore ? = 8
Finally, f(7) = 8.
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